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3 years ago

13

Why the weight of a body decreases with increases in distance from the earth's surface

2 answers:

Tom [10]3 years ago

6 0

because as the distance increases the gravitational force decreases so the weight of a body decreases

Bess [88]3 years ago

5 0

Answer:

Weight of the body decreases with increase in distance from the Earth because weight is inversely proportional to distance from the Earth which means Weight of the body decreases with increase in distance from the Earth and vice versa. Also Weight is "gravitational force exerted on the body" so as the distance from the Earth increases , the "gravitational force" decreases.

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A radio wave transmits 38.5 W/m2 of power per unit area. A flat surface of area A is perpendicular to the direction of propagati

Margaret [11]

Answer:

$P=2.57\times 10^{-7}\ N/m^2$

Explanation:

Given that,

A radio wave transmits 38.5 W/m² of power per unit area.

A flat surface of area A is perpendicular to the direction of propagation of the wave.

We need to find the radiation pressure on it. It is given by the formula as follows :

$P=\dfrac{2I}{c}$

Where

c is speed of light

Putting all the values, we get :

$P=\dfrac{2\times 38.5}{3\times 10^8}\\\\=2.57\times 10^{-7}\ N/m^2$

So, the radiation pressure is $2.57\times 10^{-7}\ N/m^2$ .

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3 years ago

When the spring, with the attached 275.0 g mass, is displaced from its new equilibrium position, it undergoes SHM. Calculate the

topjm [15]

Answer:

The period of oscillation is 1.33 sec.

Explanation:

Given that,

Mass = 275.0 g

Suppose value of spring constant is 6.2 N/m.

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Using formula of angular frequency

$\omega=\sqrt{\dfrac{k}{m}}$

Where, m = mass

k = spring constant

Put the value into the formula

$\omega=\sqrt{\dfrac{6.2}{275.0\times10^{-3}}}$

$\omega=4.74\ rad/s$

We need to calculate the period of oscillation,

Using formula of time period

$T=\dfrac{2\pi}{\omega}$

Put the value into the formula

$T=\dfrac{2\pi}{4.74}$

$T=1.33\ sec$

Hence, The period of oscillation is 1.33 sec.

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2 years ago

What is the special name for doctors who deal with cancer?

Sladkaya [172]

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3 years ago

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nirvana33 [79]

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3 years ago

Suppose the odometer on our car is broken and we want to estimate the distance driven over a 30 second time interval. We take th

Natalija [7]

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so here we will have

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4 0

3 years ago

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